Question

An ideal gas in thermally insulated vessel at internal pressure $=p_1,volume=V_1$ and absolute temperature $=T_1$ expands irreversibly against zero external pressure, as shown in the diagram.
The final internal pressure, volume and absolute temperature of the gas are $p_2,V_{2}and{T}_2,$ respectively. For this expansion

• A. q = 0
• B. $T_2=T_1$
• C. $p_2{V}_2=p_1{V}_1$
• D. $p_2{V}_2^{\gamma }=p_1{V}_1^{\gamma }$

Answer 1

Answer: (A), (B), (C)

$p_2{V}_2=p_1{V}_1$

PLAN This problem includes concept of isothermal adiabatic irreversible expansion.
Process is adiabatic because of the use of thermal insolution therefodre, q = 0
$\because P_{ext}=0$
$w=P_{ext}.△V=0\times △V=0$
Internal energy can be written as \triangle U = q + W = 0
The change in internal energy fo an ideal gas depends only on temperature and change in internal energy ($△$ U) = 0 therefore, $△$ T = 0 hence, process is isothermal and $T_2=T_{1}and{p}_2{V}_2=p_1{V}_{11}$
(d) $p_2{V}_2^{\gamma }=\left(d\right)p_1{V}_1^{\gamma }$ is incorrect, it is valid for adiabatic reversible process.
Hence, only (a), (b) and (c) are correct choices.